PS and PT are tangents drawn to the circle with centre O.
Angle SOT is 120° .
Prove that 4PS square = 3OP square.
Please solve this without trignometry ..With Pythagoras theoremo or any other
Answers
Answer: This question can be solved by trigonometry ONLY.
Step-by-step explanation:
Given : PT and PS are tangent of the circle.
O is the centre of the circle and ∠SPT = 120°
To prove : OP = 2 PS.
Proof : In ΔOPS and ΔOPT
∠OSP = ∠OTP (each 90° bcz radius is perpendicular to the tangent)
PS = PT (tangent from an external point are equal)
OS = OT (radius of the circle are equal)
So,
ΔOPS ≅ ΔOPT (by SAS similarity criterion)
Therefore,
∠OPS = ∠OPT (by CPCT)
∠POS = ∠POT (by CPCT)
As,
∠SPT = 120° and ∠SOT = 60°
So, now;
∠OPS = ∠OPT = 60°
∠POS = ∠POT = 30°
Now, by taking ΔPOS
We get,
=> sin30° = PS/PO
=> 1/2 = PS/PO
=> PO = 2PS
Hence Proved.
If you did not understand this, let me know. I will explain
Where the two tangents PS and PT are drawn to the circle .
<SOT =120°
then OS and OT are the radii of the circle
so OS = OT
<OSP= <OTP = 90°
∆ SPO ≈ ∆PTO by SAS congruency
Thus, <POT = < POS = 120°/2= 60°
So
<POT+ <OTP+<OPT= 180°
60°+ 90°+x = 180°
<OPT = 180°-150°
<OPT = 30°
Where <OPT = 30° = <OPS
Then
In Pythagoras theorem
Op²= ps²+os²
So 4PS = 3Op